Problem 4

Since $0<x<45^\circ$, we have $t=\tan x\in(0,1)$, so the factor $t^2$ contributes no admissible root. The factor $t^2-4t+1=0$ gives $t=2\pm\sqrt3$, and the root in $(0,1)$ is $t=2-\sqrt3=\tan15^\circ$, so $x=15^\circ$. The factor $t^2+2t-1=0$ gives $t=-1\pm\sqrt2$, and the root in $(0,1)$ is $t=\sqrt2-1=\tan22.5^\circ$, so $x=22.5^\circ$. Both values indeed satisfy $\angle BEK=45^\circ$ exactly (direct substitution back into the coordinates of Steps 1–3 confirms $\vec{EB}\cdot\vec{EK}=\tfrac{\sqrt2}{2}\,|\vec{EB}|\,|\vec{EK}|$ at each), and $0<15^\circ,22.5^\circ<45^\circ$. Since $\angle CAB=4x$, the possible values are $\angle CAB=60^\circ$ and $\angle CAB=90^\circ$. $\blacksquare$